A wire is bent into three segments each of radius r. Each segment is a quadrant of a circle. The first segment lies in X-Y plane, the second in the Y-Z plane and third in X-Z plane. If a current i flows in the wire, then net dipole moment of wire is

The circular segments in each of the quadrants can be considered individually for calculating the magnetic moment. Thus, segments in each of the quadrants are considered as loops individual by joining the two ends hypothetically in the same plane. When all the three segments are considered together, the contribution by hypothetical elements are cancelled.

$\begin{array}{l}{\mathrm{M}}_1=\mathrm{i}\left(\frac{{\mathrm{\pi a}}^2}{4}\right)\widehat{\mathrm{k}}\\ \mathrm{Similarly},{\mathrm{M}}_2=\mathrm{i}\left(\frac{{\mathrm{\pi a}}^2}{4}\right)\widehat{\mathrm{j}}\\ {\mathrm{M}}_3=\mathrm{i}\left(\frac{{\mathrm{\pi a}}^2}{4}\right)\widehat{\mathrm{i}}\end{array}$

So, net magnetic moment

$\begin{array}{l}\overrightarrow{\mathrm{M}}={\overrightarrow{\mathrm{M}}}_1+{\overrightarrow{\mathrm{M}}}_2+{\overrightarrow{\mathrm{M}}}_3\\ \overrightarrow{\mathrm{M}}=\frac{{\mathrm{\pi a}}^2\mathrm{i}}{4}\left(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right)\end{array}$